11:45 no, that would just be z=-1 in our branch. in fact, the change z->-z-3 does not change the formula for y, meaning it is a bijection between branches (every solution can be presented as a solution of whichever branch you wish)
@@gianluca4893 in the case when it is a change of variables that doesn't change the equations (that includes, for example, the fact that z is still an integer), then yes. In our case, you could say (if you want to be more rigorous) "for the solutions of the second branch, let z=-3-z_1, then z_1 is an integer, y=(-3-z_1)(-3-z_1+3)=z_1(z_1+3) and x³=-(-3-z_1)(-3-z_1+2)(-3-z_1+3)=z_1(z_1+1)(z_1+3). That is the form of a solution of the first branch with z=z_1, meaning we already covered that case." Hope this helped!
@@bot24032 Yeah, at this point we're assuming x=z, so if you start off by plugging in x=0 to get y=-2, the z value derived from that is no longer equal to x, so no solutions were missed as it's a bijection as you say
@@gianluca4893 I was competing in some smaller olympiads and was planning to do the national one, but have to move countries and to compete now I would have to a) know German well enough to write solutions in it and b) have a good understanding of what is and isn't allowed to be used without proof in my solutions (I don't even know the school curriculum, so I'm never sure if I even can use what I think of as basic)
considering the given eqn as quadratic in x^3, and applying the condition that the discriminant is a perfect square,we get 4y+9 = k^2 for some k€ N. putting this value of y in the given eqn and solving for x^3, we get -x^3= (k^2-9)(1+ k)/8 or -x^3 =(k^2-9)(1-k)/8 thus, we must have (k^2-9)(1+k) is a perfect cube or (k^2-9)(1-k) is a perfect cube in case 1, if (k^2-9)(1+k) =c^3 for some integer c. clearly,k =3 gives c=0 as a solution for values of k>9, we get k^2- 9k -9 =(c-k)(c^2 +ck + k^2) for values of k>9, we have l.hs of above eqn and hence, r.hs are both positive integers. thus, 0
I'm afraid we have to consider x = z+1, particularly as it leads to the solution x=2, y=4. We are not looking at linear relationships, but actually at quadratic ones, so as x increases from large and negative to large and positive, the value of the expression (x^3)^2 + y(x^3) - (y^3 + 2y^2) decreases from large and positive, through a minimum, and increases back to large and positive again. The value of y determines if the expression is ever negative, and therefore would have an opportunity to attain a value of 0, but that might occur twice, before or after the minimum. You'll find that y has to be greater than -3 for the possibility of the expression reaching zero to exist. Trichotomy isn't at play here.
@@hagenfarrell Are these 10 colors standard for colored chalk? I only ask because I bought some pretty good colored chalk (not hagoromo, unfortunately) and it's the *exact* same colors
Review the video again from 2:45. 4y+9 must be equal to t^2 where t is in an integer (so that the square root yields an integer). Such a t must be odd 'cause 4y+9 is odd, so we might as well write t=2z+3 where z is an integer.
11:45 no, that would just be z=-1 in our branch. in fact, the change z->-z-3 does not change the formula for y, meaning it is a bijection between branches (every solution can be presented as a solution of whichever branch you wish)
If there Is a bijection then the solutions Will be the same?
@@gianluca4893 in the case when it is a change of variables that doesn't change the equations (that includes, for example, the fact that z is still an integer), then yes. In our case, you could say (if you want to be more rigorous) "for the solutions of the second branch, let z=-3-z_1, then z_1 is an integer, y=(-3-z_1)(-3-z_1+3)=z_1(z_1+3) and x³=-(-3-z_1)(-3-z_1+2)(-3-z_1+3)=z_1(z_1+1)(z_1+3). That is the form of a solution of the first branch with z=z_1, meaning we already covered that case." Hope this helped!
@@bot24032 Yeah, at this point we're assuming x=z, so if you start off by plugging in x=0 to get y=-2, the z value derived from that is no longer equal to x, so no solutions were missed as it's a bijection as you say
@@bot24032 Wow, thank you so much! Great explanation! Do you compete in math olympiad?
@@gianluca4893 I was competing in some smaller olympiads and was planning to do the national one, but have to move countries and to compete now I would have to a) know German well enough to write solutions in it and b) have a good understanding of what is and isn't allowed to be used without proof in my solutions (I don't even know the school curriculum, so I'm never sure if I even can use what I think of as basic)
considering the given eqn as quadratic in x^3, and applying the condition that the discriminant is a perfect square,we get 4y+9 = k^2 for some k€ N.
putting this value of y in the given eqn and solving for x^3, we get
-x^3= (k^2-9)(1+ k)/8
or -x^3 =(k^2-9)(1-k)/8
thus, we must have
(k^2-9)(1+k) is a perfect cube or (k^2-9)(1-k) is a perfect cube
in case 1, if (k^2-9)(1+k) =c^3 for some integer c.
clearly,k =3 gives c=0 as a solution
for values of k>9, we get
k^2- 9k -9 =(c-k)(c^2 +ck + k^2)
for values of k>9, we have
l.hs of above eqn and hence, r.hs are both positive integers.
thus,
0
at 15:26 the quadratic factor was y^2 +6y+16, not y^2 +6y^2+16
I love these kind of problems
13:07 by the trichotomy property of integers, we are done. There is no need to consider x = z+1.
I'm afraid we have to consider x = z+1, particularly as it leads to the solution x=2, y=4. We are not looking at linear relationships, but actually at quadratic ones, so as x increases from large and negative to large and positive, the value of the expression (x^3)^2 + y(x^3) - (y^3 + 2y^2) decreases from large and positive, through a minimum, and increases back to large and positive again. The value of y determines if the expression is ever negative, and therefore would have an opportunity to attain a value of 0, but that might occur twice, before or after the minimum. You'll find that y has to be greater than -3 for the possibility of the expression reaching zero to exist. Trichotomy isn't at play here.
Huh, you must have been in the mood to pronounce 4 syllable words.
Explain more carefully why a>b, a=b, a
day 3 of asking Michael what brand chalk he uses
Hagoromo
@@soyoltoi does hagoromo sell a 10 color version?
@@soyoltoiJust like Papa Flammy?
More than likely Hagoromo, it’s what most mathematicians use; as do I. It’s wonderful chalk!
@@hagenfarrell Are these 10 colors standard for colored chalk? I only ask because I bought some pretty good colored chalk (not hagoromo, unfortunately) and it's the *exact* same colors
I mean how do you proove that there is no non integer z which yields and integer x and integer y
By the way this equation defines an elliptic curve over the real numbers!
20:07
this guy is indefatigable ...
18:11 (z+3)^3
Why my subscription to your channel suddenly disappeared ?
I subscribed again, so ok now.
18:13 I think the cube is wrong (should have been z^3 + 3z^2 + 9z +27)
No, you pick up a binomial 3 and a 3 from the constant term to give you the 9
Minor quibble, the empty set is an element of the empty set. The problem is the empty set is not in Z.
I hate these pedantic comments that add nothing. And it's even worse when they are wrong. Empty set doesn't have any elements. It's irrelevant tho.
interested to see belarus source of solution as well.
Mistake in proof? The 'z' variable is not intiger by definition. Becouse 2*z+3 = k^2 not implicated it. E.g. z=1/2 is valid not integer value.
You areing correctful
If z is not an integer but 2z+3 is then 2z+3 is even which is not true by definition
Missing derivation of the fact, that z is integer
Review the video again from 2:45.
4y+9 must be equal to t^2 where t is in an integer (so that the square root yields an integer).
Such a t must be odd 'cause 4y+9 is odd, so we might as well write t=2z+3 where z is an integer.